Решите неравенство log1/2(x+4)> -3

Я9-ый класс, поэтому решаю, как училась сама) log1/2(x+4)> -3 (1/2)^-3 > (x+4) 2^3> x+4 8> x+4 x< 4 ответ: (-беконечность; 4)

а) 8 1/3 + 6 1/2 + 3 5/6 = 8 2/6 + 6 3/6 + 3 5/6 = 17 10/6 = 17 5/3 = 18 2/3;

б) 12 3/8 + 5 1/4 + 7 1/2 = 12 3/8 + 5 2/8 + 7 4/8 = 24 9/8 = 25 1/8;

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в) 2 1/3 · 3/10 : 2 4/5 = 1/4

1) 2 1/3 · 3/10 = 7/3 ·3/10 = 7/10

2) 7/10 : 2 4/5 = 7/10 : 14/5 = 7/10 · 5/14 = (1·1)/(2·2) = 1/4

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г) 1 1/6 : 2 1/6 · 26 = 14

1) 1 1/6 : 2 1/6 = 7/6 : 13/6 = 7/6 · 6/13 = 7/13

2) 7/13 · 26/1 = (7·2)/(1·1) = 14

а) 3 3/5 · 1/3 + 6 4/9 : 2 = 4 целых 19/45

1) 3 3/5 · 1/3 = 18/5 · 1/3 = (6·1)/(5·1) = 6/5 = 1 1/5

2) 6 4/9 : 2 = 58/9 · 1/2 = (29·1)/(9·1) = 29/9 = 3 2/9

3) 1 1/5 + 3 2/9 = 1 9/45 + 3 10/45 = 4 19/45

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б) 2/3 - 8/23 · (3/4 + 1 1/6) = 0

1) 3/4 + 1 1/6 = 9/12 + 1 2/12 = 1 11/12 = 23/12

2) 8/23 · 23/12 = (2·1)/(1·3) = 2/3

3) 2/3 - 2/3 = 0

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в) 2 5/6 - 1 1/5 · 1 1/9 + 1 5/7 : 1 1/7 = 3

1) 1 1/5 · 1 1/9 = 6/5 · 10/9 = (2·2)/(1·3) = 4/3 = 1 1/3

2) 1 5/7 : 1 1/7 = 12/7 : 8/7 = 12/7 · 7/8 = (3·1)/(1·2) = 3/2 = 1 1/2

3) 2 5/6 - 1 1/3 = 2 5/6 - 1 2/6 = 1 3/6 = 1 1/2

4) 1 1/2 + 1 1/2 = 2 2/2 = 3

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г) 5 2/9 : (3 - 1 1/9 · 2 2/5) + 4/5 = 16 целых 7/15

1) 1 1/9 · 2 2/5 = 10/9 · 12/5 = (2·4)/(3·1) = 8/3 = 2 2/3

2) 3 - 2 2/3 = 2 3/3 - 2 2/3 = 1/3

3) 5 2/9 : 1/3 = 47/9 · 3/1 = (47·1)/(3·1) = 47/3 = 15 2/3

4) 15 2/3 + 4/5 = 15 10/15 + 12/15 = 15 22/15 = 16 7/15